Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.
AB = 2CD
We have to find the ratio of the areas of triangles AOB and COD.
From ΔAOB and ΔCOD, we have
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
∠5 = ∠6 (Vertically opposite angle)
∴ ΔAOB ~ ΔCOD [AAA similarity criterion]
As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,
Area of (ΔAOB)/Area of (ΔCOD) = AB2/CD2
= (2CD)2/CD2 [∴ AB = 2CD]
∴ Area of (ΔAOB)/Area of (ΔCOD)
= 4CD2/CD2 = 4/1
Therefore, the required ratio of the area of ΔAOB and ΔCOD = 4:1